标签：copy find 一个 思路 only The pos max geo
There is an integer sequence of length 2N: A0,A1,…,A2N?1. (Note that the sequence is 0-indexed.)
For every integer K satisfying 1≤K≤2N?1, solve the following problem:
Input is given from Standard Input in the following format:
N A0 A1 … A2N?1
Print 2N?1 lines. In the i-th line, print the answer of the problem above for K=i.
2 1 2 3 1
3 4 5
For K=1, the only possible pair of i and j is (i,j)=(0,1), so the answer is A0+A1=1+2=3.
For K=2, the possible pairs of i and j are (i,j)=(0,1),(0,2). When (i,j)=(0,2), Ai+Aj=1+3=4. This is the maximum value, so the answer is 4.
For K=3, the possible pairs of i and j are (i,j)=(0,1),(0,2),(0,3),(1,2),(1,3),(2,3) . When (i,j)=(1,2), Ai+Aj=2+3=5. This is the maximum value, so the answer is 5.
3 10 71 84 33 6 47 23 25
81 94 155 155 155 155 155
4 75 26 45 72 81 47 97 97 2 2 25 82 84 17 56 32
101 120 147 156 156 178 194 194 194 194 194 194 194 194 194
题意 ： 给你 一堆数，并设定一个变量 k，要求 k 是从 1 开始递增的，每次从不超过 k 的序列中位置上取两个数 i, j， 且要求 i or j <= k
思路分析：对于一个k, 寻找小于等于 k 中
代码示例：
const int maxn = 1e6+5;
int n;
int pre[maxn];
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
cin >> n;
for(int i =0; i < (1<<n); i++){
scanf("%d", &pre[i]);
}
int ans = 0;
for(int i = 1; i < (1<<n); i++){
int s1 = pre[0], s2 = 0;
for(int j = i; j; j = (j-1)&i){
if (pre[j] > s1) {s2 = s1, s1 = pre[j];}
else if (pre[j] > s2) s2 = pre[j];
}
ans = max(ans, s1+s2);
printf("%d\n", ans);
}
return 0;
}
标签：copy find 一个 思路 only The pos max geo
原文地址：https://www.cnblogs.com/ccut-ry/p/9265378.html
